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Ôi, trang wed không tự nhận diện được công thức latex. Mình đăng lại bài giải:
a) Ta có
\(4T=\frac{4}{1+\sqrt{5}}+\frac{4}{\sqrt{5}+\sqrt{9}}+...+\frac{4}{\sqrt{2013}+\sqrt{2017}}\)
\(=\frac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}{\sqrt{5}+1}+...+\frac{\left(\sqrt{2017}+\sqrt{2013}\right)\left(\sqrt{2017}-\sqrt{2013}\right)}{\sqrt{2017}+\sqrt{2013}}\)
\(=\sqrt{5}-1+\sqrt{9}-\sqrt{5}+\sqrt{13}-\sqrt{9}+...+\sqrt{2017}-\sqrt{2013}\)
\(=\sqrt{2017}-1\)
\(\Rightarrow T=\frac{\sqrt{2017}-1}{4}\)
b) Ta có
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{2-1}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}\sqrt{1}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
Tương tự ta có
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
......................
\(\frac{1}{100\sqrt{99}+99\sqrt{100}}=\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
Suy ra
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
a)\[\begin{array}{l}
4T = \frac{4}{{1 + \sqrt 5 }} + \frac{4}{{\sqrt 5 + \sqrt 9 }} + ... + \frac{4}{{\sqrt {2013} + \sqrt {2017} }}\\
= \frac{{(\sqrt 5 + 1)(\sqrt 5 - 1)}}{{1 + \sqrt 5 }} + ... + \frac{{(\sqrt {2017} + \sqrt {2013} )(\sqrt {2017} - \sqrt {2013} )}}{{\sqrt {2013} + \sqrt {2017} }}\\
= \sqrt 5 - 1 + \sqrt 9 - \sqrt 5 + ... + \sqrt {2017} - \sqrt {2013} \\
= 1 + \sqrt 5 - \sqrt 5 + \sqrt 9 - \sqrt 9 + ... + \sqrt {2013} - \sqrt {2013} + \sqrt {2017} \\
= 1 + \sqrt {2017} \\
\Rightarrow T = \frac{{1 + \sqrt {2017} }}{4}
\end{array}\]
\(E=\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}.\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)\)
\(E=\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{5\sqrt{6}-12}{18\sqrt{2}}\)
\(E=\frac{36\sqrt{2}}{18\sqrt{6}}+\frac{12\sqrt{3}}{18\sqrt{6}}+\frac{\left(5\sqrt{6}-12\right).\sqrt{3}}{18\sqrt{3}}\)
\(E=\frac{36\sqrt{2}+12\sqrt{3}+\left(5\sqrt{6}-12\right).\sqrt{3}}{18\sqrt{6}}\)
\(E=\frac{51\sqrt{2}}{18\sqrt{6}}\)
\(E=\frac{17\sqrt{2}}{6\sqrt{6}}\)
\(E=\frac{17\sqrt{2}}{2.3\sqrt{2}.\sqrt{3}}\)
\(E=\frac{17}{\sqrt{2}.3\sqrt{2}.\sqrt{3}}\)
\(E=\frac{17}{6\sqrt{3}}\)
\(E=\frac{17\sqrt{3}}{18}\)
\(a,\frac{2}{3+2\sqrt{2}}-\frac{7}{1-2\sqrt{2}}+\frac{4}{\sqrt{5}-1}+\sqrt{8}-2\)
\(=\frac{2.\left(3-2\sqrt{2}\right)}{9-8}-\frac{7.\left(1+2\sqrt{2}\right)}{1-8}+\frac{4.\left(\sqrt{5}+1\right)}{5-1}+2\sqrt{2}-2\)
\(=6-4\sqrt{2}-\frac{7.\left(1+2\sqrt{2}\right)}{-7}+\frac{4.\left(\sqrt{5}+1\right)}{4}+2\sqrt{2}-2\)
\(=6-4\sqrt{2}+1+2\sqrt{2}+\sqrt{5}+1+2\sqrt{2}-2\)
\(=6+\sqrt{5}\)
\(b,\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{5}}\)
\(=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{5}}{4-5}\)
\(=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{5}}{-1}\)
\(=-1+\sqrt{2}+\sqrt{3}-\sqrt{2}-2+\sqrt{5}\)
\(=-3+\sqrt{3}+\sqrt{5}\)
\(c,\sqrt{4-2\sqrt{3}}+2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{3}\)
\(=\sqrt{3}-1+2\sqrt{3}\)
\(=-1+3\sqrt{3}\)
\(d,A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{2\sqrt{3}}{\sqrt{2}}\)
\(=\sqrt{6}\)
\(e,B=\sqrt{\frac{2}{2+\sqrt{3}}}\)
Ta có \(\frac{2}{2+\sqrt{3}}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}\)
Thay lại ta được \(\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
.... Đúng thì ủng hộ nha ....
Kết bạn với mình ... ;) ;)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{\sqrt{3}\left(2+\sqrt{6}\right)+\sqrt{3}\left(2-\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}-\frac{1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(-2\sqrt{3}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{1}{\sqrt{2}}-\frac{2-\sqrt{6}}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}+\frac{\left(\sqrt{2}-1\right)\left(-2\sqrt{6}+6\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{1}{\sqrt{2}}\)
\(=\frac{2-\sqrt{6}}{2}-4\sqrt{3}+6\sqrt{2}+2\sqrt{6}-6\)
\(=6\sqrt{2}-4\sqrt{3}+\frac{3\sqrt{6}}{2}-5\)
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